Hello:
the equation is : y = ax+b
the slope is a : a×(5) = -1......( perpendicular to a line with a slope of 5)
a = -1/5 y=(-1/5)x+b
the line that passes through (-3, 2) 2 = (-1/5)(-3)+b
b =7/5
the equation is : y = (-1/5)x+7/5
Answer:
[tex]y=-\frac{1}{5}x+\frac{7}{5}[/tex]
Step-by-step explanation:
Line LM [tex]y=5x+4[/tex]
Find the equation of the line in slope intercept form perpendicular to LM
In y=mx+b , the slope is m
Line LM [tex]y=5x+4[/tex]
Slope of the given line is 5.
Slope of the perpendicular line is the negative reciprocal of the slope of the given line
Slope of the perpendicular line is [tex]-\frac{1}{5}[/tex]
Use point slope form
[tex]y-y_1=m(x-x_1)[/tex]
(x1,y1) is (-3,2)
[tex]y-2=-\frac{1}{5}(x-(-3))[/tex]
[tex]y-2=-\frac{1}{5}(x+3)[/tex]
[tex]y-2=-\frac{1}{5}x-\frac{3}{5}[/tex]
Add 2 on both sides
[tex]y=-\frac{1}{5}x-\frac{3}{5}+2[/tex]
[tex]y=-\frac{1}{5}x+\frac{7}{5}[/tex]
